一道关于三角函数的数学题设tan(A+8/7π)=a . 求证:[sin(15/7π+A)+3cos（A-13/7π)]

设x=α+8π/7,则有:tanx=a
∴15π/7+α=π+（α+8π/7）=π+x
α-13π/7=(α+8π/7)-3π=x-3π
20π/7-α=4π-(α+8π/7)=4π-x
α+22π/7=(α+8π/7)+2π=x+2π
于是,原所求证等式左侧:
左侧=[sin(π+x)+3cos(x-3π)]/[sin(4π-x)-cos(x+2π)]
=(-sinx-3cosx)/(-sinx-cosx)
=(sinx+3cosx)/(sinx+cosx)
=[(sinx+3cosx)/cosx]/[(sinx+cosx)/cosx]
=(tanx+3)/(tanx+1)
=(a+3)/(a+1)

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